1. Chapter 3 Euclidean Vector Spaces

2. Outline 3.1 Vectors in 2-Space, 3-Space, and n-Space
3.2 Norm, Dot Product, and Distance in Rn
3.3 Orthogonality
3.4 The Geometry of Linear Systems
3.5 Cross Product

3. 3.1 Vectors in 2-Space, 3-Space, and n-Space

4. Geometric Vectors
In this text, vectors are denoted in bold face type such as a, b, v, and scalars are denoted in lowercase italic type such as a, b, v.
A vector v has initial point A and terminal point B
Vectors with the same length and direction are said equivalent.
The vector whose initial and terminal points coincide has length zero, and is called zero vector, denoted by 0. B A

5. Definitions
If v and w are any two vectors, then the sum v + w is the vector determined as follows:
Position the vector w so that its initial point coincides with the terminal point of v. The vector v + w is represented by the arrow from the initial point of v to the terminal point of w.
If v and w are any two vectors, then the difference of w from v is defined by v – w = v + (-w).
If v is a nonzero vector and k is nonzero real number (scalar), then the product kv is defined to be the vector whose length is |k| times the length of v and whose direction is the same as that of v if k > 0 and opposite to that of v if k < 0. We define kv = 0 if k = 0 or v = 0.
A vector of the form kv is called a scalar multiple.

6. Examples

7. Vectors in Coordinate Systems

8. Vectors in 3-Space
and are equivalent if and only if v1=w1, v2=w2, v3=w3

9. Vectors
If the vector has initial point P1 (x1, y1, z1) and terminal point P2 (x2, y2, z2), then z y x

10. Theorem 3.1.1 (Properties of Vector Arithmetic)
If u, v and w are vectors in Rn and k and l are scalars, then the following relationships hold.
u + v = v + u
(u + v) + w = u + (v + w)
u + 0 = 0 + u = u
u + (-u) = 0
k(lu) = (kl)u
k(u + v) = ku + kv
(k + l)u = ku + lu
1u = u

11. Proof of part (b) (geometric)

12. Theorem and Definition
Theorem 3.1.2: If v is a vector in Rn and k is a scalar, then:
0v = 0
k0 = 0
(-1)v = -v
If w is a vector in Rn, then w is said to be a linear combination of the vectors v1, v2, …, vr in Rn if it can be expressed in the form
where k1, k2, …, kr are scalars.

13. Alternative Notations for Vectors
Comma-delimited form:
It can also written as a row-matrix form
Or a column-matrix form

14. 3.2 Norm, Dot Product, and Distance in Rn

15. Norm of a Vector
The length of a vector u is often called the norm (範數) or magnitude of u and is denoted by ||u||.
It follows from the Theorem of Pythagoras (畢達哥拉斯) that the norm of a vector u = (u1,u2,u3) in 3-space is
P(u1,u2,u3) O S Q R

16. Norm of a Vector
If v=(v1, v2, …, vn) is a vector in Rn, then the norm of v is denoted by ||v||, and is defined by
Example:
The norm of v=(-3,2,1) in R3 is
The norm of v=(2, -1, 3, -5) in R4 is

17. Theorem 3.2.1 If v is a vector in Rn, and if k is any scalar, then:
||v|| = 0 if and only if v=0
||kv|| = |k| ||v||
Proof of (c):
If v = (v1, v2, …, vn), then kv = (kv1, kv2, …, kvn), so

18. Unit Vector A vector of norm 1 is called a unit vector. (單位向量)
You can obtain a unit vector in a desired direction by choosing any nonzero vector v in that direction and multiplying v by the reciprocal of its length.
The process is called normalizing v
Example: v = (2,2,-1),
You can verify that

19. Standard Unit Vectors
When a rectangular coordinate system is introduced in R2 or R3, the unit vectors in the positive directions of the coordinates axes are called standard unit vectors.
In R2, i = (1,0) and j = (0,1)
In R3, i = (1,0, 0), j = (0, 1, 0), k = (0, 0, 1)
Every vector v=(v1,v2) in R2 can be expressed as a linear combination of standard unit vectors
(0,1) j i
(1,0)
(0,0,1) k j i
(0,1,0)
(1,0,0)

20. Standard Unit Vectors
We can generalize these formulas to Rn by defining standard unit vectors in Rn to be
Every vector v=(v1,v2,…,vn) in Rn can be expressed as
Example: (2,-3,4) = 2i – 3j + 4k
(7,3,-4,5) = 7e1 + 3e2 – 4e3 + 5e4 …

21. Distance The distance between two points is the norm of the vector.
If P1(x1, y1, z1) and P2(x2, y2, z2) are two points in 3-space, then the distance d between them is the norm of the vector
Euclidean distance (歐幾里德距離, 歐式距離)
If u = (u1, u2,…, un) and v=(v1, v2, …, vn) are points in Rn, then the distance d(u,v) is defined as

22. Definitions
Let u and v be two nonzero vectors in 2-space or 3-space, and assume these vectors have been positioned so their initial points coincided. By the angle between u and v, we shall mean the angle  determined by u and v that satisfies 0    .
If u and v are vectors in 2-space or 3-space and  is the angle between u and v, then the dot product (點積) or Euclidean inner product (內積) u · v is defined by

23. Dot Product
If the vectors u and v are nonzero and  is the angle between them, then
 is acute (銳角) if and only if u · v > 0
 is obtuse (鈍角) if and only if u · v < 0
 = /2 (直角) if and only if u · v = 0

24. Example
If the angle between the vectors u = (0,0,1) and v = (0,2,2) is 45, then
Component Form
of Dot Product

25. Example
Find the angle between a diagonal of a cube and one of its edges
(0,0,k) u3
(k,k,k) d u2
(0,k,0) u1
(k,0,0)

26. Component Form of Dot Product
Let u=(u1,u2,u3) and v=(v1,v2,v3) be two nonzero vectors.
According to the law of cosine P Q
law of cosine

27. Component Form of Dot ProductQ

28. Definition
If u=(u1,u2,…,un) and v=(v1,v2,…,vn) are vectors in Rn, then the dot product (點積) (also called the Euclidean inner product (內積)) of u and v is denoted by u．v and is defined by
Example: u=(-1,3,5,7) and v=(-3,-4,1,0)
u．v = (-1)(-3) + (3)(-4) + (5)(1) + (7)(0) = -4

29. Theorems The special case u = v, we obtain the relationship
If u, v and w are vectors in 2- or 3-space, and k is a scalar, then
u · v = v · u [symmetry property]
u · (v + w) = u · v + u · w [distributive property]
k(u · v) = (ku) · v = u · (kv) [homogeneity property]
v · v ≧ 0 and v · v = 0 if v = 0 [positivity property]

30. Proof of Theorem 3.2.2 k(u · v) = (ku) · v = u · (kv)
Let u=(u1,u2,u3) and v=(v1,v2,v3)

31. Theorem 3.2.3
If u, v, and w are vectors in Rn, and if k is a scalar, then
0．v = v．0 = 0
(u+v)．w = u．w + v．w
u．(v-w) = u．v – u．w
(u-v)．w = u．w - v．w
k(u．v) = u．(kv)
Proof(b)
[by symmetry]
[by distributivity]
[by symmetry]

32. Example Calculating with dot products
(u - 2v)．(3u + 4v) =u．(3u + 4v) – 2v．(3u + 4v) =3(u．u) + 4(u．v) – 6(v．u) – 8(v．v) =3||u||2 – 2(u．v) – 8||v||2

33. Cauchy-Schwarz Inequality
With the formula
The inverse cosine is not defined unless its argument satisfies the inequalities
Fortunately, these inequalities do hold for all nonzero vectors in Rn as a result of Cauchy-Schwarz inequality

34. Theorem 3.2.4 Cauchy-Schwarz Inequality
If u = (u1,u2,…,un) and v=(v1,v2,…,vn) are vectors in Rn, then |u．v| ≦||u|| ||v|| or in terms of components
We will omit the proof of this theorem because later in the
text we will prove a more general version of which this will
be a special case.

35. To show
Cauchy-Schwarz Inequality: If u = (u1,u2,…,un) and v=(v1,v2,…,vn) are vectors in Rn, then |u．v| ≦||u|| ||v||

36. Geometry in Rn
The sum of the lengths of two side of a triangle is at least as large as the third
The shortest distance between two points is a straight line
Theorem 3.2.5
If u, v, and w are vectors in Rn, and k is any scalar, then
||u+v|| ≦ ||u|| + ||v||
d(u,v) ≦ d(u,w) + d(w,v)
u+v v u

37. Proof of Theorem 3.2.5 Proof (a) Proof (b) Property of absolute value
Cauchy-Schwarz inequality
based on (a)

38. Theorem 3.2.6 Parallelogram Equation for Vectors
If u and v are vectors in Rn, then ||u+v||2 + ||u-v||2 = 2(||u||2 + ||v||2)
Proof:
u-v v
u+v u

39. Theorem 3.2.7
If u and v are vectors in Rn with the Euclidean inner product, then
Proof:

40. Dot Products as Matrix Multiplication
View u and v as column matrices
Example:

41. Dot Products as Matrix Multiplication
If A is an n×n matrix and u and v are n×1 matrices
You can check

42. Dot Product View of Matrix Multiplication
If A=[aij] is a m×r matrix, and B=[bij] is an r×n matrix, then the ijth entry of AB is
which is the dot product of the ith row vector of A
and the jth column vector of B

43. Dot Product View of Matrix Multiplication
If the row vectors of A are r1, r2, …, rm and the column vectors of B are c1, c2, …, cn, then the matrix product AB can be expressed as

44. 3.3 Orthogonality

45. Orthogonal Vectors Recall that It follows that if and only if u．v = 0
Definition: Two nonzero vectors u and v in Rn are said to be orthogonal [正交] (or perpendicular [垂直]) if u．v = 0.
The zero vector in Rn is orthogonal to every vector in Rn.
A nonempty set of vectors in Rn is called an orthogonal set if all pairs of distinct vectors in the set are orthogonal.
An orthogonal set of unit vectors is called an orthonormal set.

46. Example Show that u=(-2,3,1,4) and v=(1,2,0,-1) are orthogonal
Show that the set S={i,j,k} of standard unit vectors is an orthogonal set in R3
We must show

47. Normal
One way of specifying slope and inclination is the use a nonzero vector n, called normal (法向量) that is orthogonal to the line or plane.
The line through the point (x0,y0) has normal n=(a,b) y
Example: the equation 6(x-3) + (y+7) = 0 represents the line through (3,-7) with normal n=(6,1)
P(x,y)
(a,b) n
P0(x0,y0) x

48. Theorem 3.3.1
If a and b are constants that are not both zero, then an equation of the form ax+by+c = 0 represents a line in R2 with normal n=(a,b)
If a, b, and c are constants that are not all zero, then an equation of the form ax+by+cz+d = 0 represents a line in R3 with normal n=(a,b,c)

49. Example
The equation ax+by=0 represents a line through the origin in R2. Show that the vector n=(a,b) is orthogonal to the line, that is, orthogonal to every vector along the line.
Solution:
Rewrite the equation as Therefore, the vector n is orthogonal to every vector (x,y) on the line.

50. An Orthogonal Projection
To "decompose" a vector u into a sum of two terms, one parallel to a specified nonzero vector a and the other perpendicular to a.
We have w2 = u – w1 and w1 + w2 = w1 + (u – w1) = u
The vector w1 is called the orthogonal projection (正交投影) of u on a or sometimes the vector component (分向量) of u along a, and denoted by projau
The vector w2 is called the vector component of u orthogonal to a, and denoted by w2 = u – projau

51. Theorem 3.3.2 Projection Theorem
If u and a are vectors in Rn, and if a≠0, then u can be expressed in exactly one way in the form u=w1+w2, where w1 is a scalar multiple of a and w2 is orthogonal to a.
Proof:
Since w1 is to be a scalar multiple of a, it has the form: w1 = ka
Our goal is to find a value of k and a vector w2 that is orthogonal to a such that u=w1+w2.
Rewrite u=w1+w2=ka+w2 , and then applying Theorems and to obtain u．a=(ka+w2)．a=k||a||2+(w2．a)
Since w2 is orthogonal to a, u．a = k||a||2 , from which we obtain
Therefore, we can get

52. Projection Theorem
The vector w1 is called the orthogonal projection of u on a, or the vector component of u along a.
The vector w2 is called the vector component of u orthogonal to a.

53. Example L 1
Find the orthogonal projections of the vectors e1=(1,0) and e2=(0,1) on the line L that makes an angle θ with the positive x-axis in R2.
Solution:
is a unit vector along L.
Find orthogonal projection of e1 along a.

54. Example
Solution:

55. Length of Orthogonal Projection
scalar
Theorem 3.2.1
Since
If denotes the angle between u and a, then

56. Length of Orthogonal Projectionu a u a

57. Theorem 3.3.3 Theorem of Pythagoras
If u and v are orthogonal vectors in Rn with the Euclidean inner product, then
||u+v||2 = ||u||2 + ||v||2
Proof:
Since u and v are orthogonal, u．v=0, then

58. Theorem 3.3.4
(a) In R2 the distance D between the point P0(x0,y0) and the line ax+by+c=0 is
(b) In R3 the distance D between the point P0(x0,y0,z0) and the plane ax+by+cz+d = 0 is

59. Proof of Theorem 3.3.4(b)
Let Q(x1,y1,z1) be any point in the plane. Position the normal n=(a,b,c) so that its initial point is at Q.
D is the length of the orthogonal projection of on n.
P0(x0,y0,z0) D
Q(x1,y1,z1)

60. Proof of Theorem 3.3.4(b)
Since the point Q(x1,y1,z1) lies in the given plane, ax1+by1+cz1+d = 0, or d=-ax1-by1-cz1
Thus

61. Example
Find the distance D from the point (1,-2) to the line 3x+4y-6 = 0 is

62. Distance Between Parallel Plane
Two planes x+2y-2z=3 and 2x+4y-4z=7
To find the distance D between the planes, we can select an arbitrary point in one of the planes and compute its distance to the other plane.
By setting y=z=0 in the equation x+2y-2z=3, we obtain the point P0(3,0,0) in this plane.
The distance between P0 and the plane 2x+4y-4z=7 is P0 D

63. 3.4 The Geometry of Linear Systems

64. Vector and Parametric Equations
A unique line in R2 or R3 is determined by a point x0 on the line and a nonzero vector v parallel to the line
A unique plane in R3 is determined by a point x0 in the plane and two noncollinear vectors v1 and v2 parallel to the plane v1 v x0 x0 v2

65. Vector and Parametric Equations
If x is a general point on such a line, the vector x-x0 will be some scalar multiple of v
x-x0 = tv or equivalently x = x0 + tv
As the variable t (called parameter) varies from - ∞ to ∞, the point x traces out the line L. x L
x-x0 x0 v

66. Theorem 3.4.1
Let L be the line in R2 or R3 that contains the point x0 and is parallel to the nonzero vector v. Then the equation of the line through x0 that is parallel to v is
x = x0 + tv
If x0=0, then the line passes through the origin and the equation has the form
x = tv
The translation by x0 of the line through the origin x L
x-x0 x0
x = tv v

67. Vector and Parametric Equations
If x is any point in the plane, then by forming suitable scalar multiples of v1 and v2, we can create a parallelogram with diagonal x-x0 and adjacent sides t1v1 and t2v2. Thus we have
x – x0 = t1v1 + t2v2 or equivalently x = x0 + t1v1 + t2v2
As the variables t1 and t2 (parameters) vary independently from –∞ to ∞, the point x varies over the entire plane W. x
t2v2 W
t1v1 x0

68. Theorem 3.4.2
Let W be the plane in R3 that contains the point x0 and is parallel to the noncollinear vectors v1 and v2. Then an equation of the plane through x0 that is parallel to v1 and v2 is given by
x = x0 +t1v1+t2v2
If x0=0, then the plane passes through the origin and the equation has the form
x = t1v1+t2v2

69. Definition
If x0 and v are vectors in Rn, and if v is nonzero, then the equation x = x0+tv defines the line through x0 that is parallel to v. In the special case where x0 = 0, the line is said to pass through the origin.
If x0, v1 and v2 are vectors in Rn, and if v1 and v2 are not collinear, then the equation x = x0+t1v1 + t2v2 defines the plane through x0 that is parallel to v1 and v2. In the special case where x0 = 0, the line is said to pass through the origin.

70. Vector Forms
The previous equations are called vector forms of a line and plane in Rn.
If the vectors in these equations are expressed in terms of their components and the corresponding components on each side are equated, then the resulting equations are called parametric equations of the line and plane.

71. Example
Find a vector equation and parametric equations of the line in R3 that passes through the point P0(1,2,-3) and is parallel to the vector v=(4,-5,1)
Solution:
The line is x = x0 + tv
If we let x=(x,y,z), and if we take x0=(1,2,-3) then this equation is (x,y,z)= (1,2,-3) + t(4,-5,1)
Equating corresponding components on the two sides of this equation yields the parametric equations x = 1+4t, y = 2-5t, z = -3+t

72. Example Find vector and parametric equations of the plane x-y+2z = 5
Solution: solving for x in terms of y and z yields x = 5+y-2z
Then using y and z as parameters t1 and t2, respectively, yields the parametric equations:
x = 5+t1-2t2, y = t1, z=t2
To obtain a vector equation of the plane we rewrite these parametric equations as (x,y,z) = (5+t1-2t2, t1, t2), or equivalently as (x,y,z) = (5,0,0) + t1(1,1,0) + t2(-2,0,1)

73. Example
Find vector and parametric equations of the plane in R4 that passes through the point x0=(2,-1,0,3) and is parallel to both v1=(1,5,2,-4) and v2=(0,7,-8,6)
Solution: the vector equation x=x0+t1v1+t2v2 can be expressed as
(x1,x2,x3,x4) = (2,-1,0,3) + t1(1,5,2,-4) + t2(0,7,-8,6)
Which yields the parametric equations
x1 = 2+t1, x2 = -1+5t1+7t2, x3 = 2t1-8t2, x4=3-4t1+6t2

74. Lines Through Two pointsx1 v x0
If x0 and x1 are distinct points in Rn, then the line determined by these points is parallel to the vector v = x1-x0
The line can be expressed as x = x0 + t(x1-x0)
Or equivalently as x=(1-t)x0 + tx1
These are called the two-point vector equations of a line in Rn

75. Example
Find vector and parametric equations for the line in R2 that passes through the points P(0,7) and Q(5,0)
Solution: Let’s choose x0=(0,7) and x1=(5,0). x1-x0 = (5,-7) and hence (x,y) = (0,7) + t(5,-7)
We can rewrite in parametric form as x = 5t, y = 7-7t

76. Definition
If x0 and x1 are vectors in Rn, then the equation x = x0 + t(x1-x0) (0 ≦ t ≦ 1) defines the line segment from x0 to x1.
When convenient, it can be written as x=(1-t)x0 + tx1 (0 ≦ t ≦ 1)
Example: the line segment from x0=(1,-3) to x1=(5,6) can be represented by x = (1,-3) + t(4,9) (0 ≦ t ≦ 1) or x = (1-t)(1,-3) + t(5,6) (0 ≦ t ≦ 1)

77. Dot Product Form of a Linear System
Recall that a linear equation has the form
a1x1+a2x2+…+anxn = b (a1,a2, …, an not all zero)
The corresponding homogeneous equation is
a1x1+a2x2+…+anxn = 0 (a1,a2, …, an not all zero)
These equations can be rewritten in vector form by letting
a = (a1,a2,…,an) and x=(x1,x2,…,xn)
Two equations can be written as

78. Dot Product Form of a Linear System
It reveals that each solution vector x of a homogeneous equation is orthogonal to the coefficient vector a.
Consider the homogeneous system
If we denote the successive row vectors of the coefficient matrix by r1, r2, …, rm, then we can write this system as

79. Theorem 3.4.3
If A is an m × n matrix, then the solution set of the homogeneous linear system Ax=0 consists of all vectors in Rn that are orthogonal to every row vector of A.
Example: the general solution of
is x1=-3r-4s-2t, x2=r, x3=-2s, x4=s, x5=t, x6=0
Vector form: x = (-3r-4s-2t, r, -2s, s, t, 0)

80. Theorem 3.4.3
According to Theorem 3.4.3, the vector x must be orthogonal to each of the row vectors
r1 = (1,3,-2,0,2,0)
r2 = (2, 6, -5, -2, 4, -3)
r3 = (0,0,5,10,0,15)
r4 = (2,6,0,8,4,18)
Verify that r1．x = 1(-3r-4s-2t)+3(r)+(-2)(-2s)+0(s)+2(t)+0(0) = 0

81. The Relationship Between Ax=0 and Ax=b
Compare the solutions of the corresponding linear systems
Homogeneous system: x1=-3r-4s-2t, x2=r, x3=-2s, x4=s, x5=t, x6=0
Nonhomogeneous system: x1=-3r-4s-2t, x2=r, x3=-2s, x4=s, x5=t, x6=1/3

82. The Relationship Between Ax=0 and Ax=b
We can rewrite them in vector form:
Homogeneous system: x = (-3r-4s-2t, r, -2s, s, t, 0)
Nonhomogeneous system: x = (-3r-4s-2t, r, -2s, s, t, 1/3)
By splitting the vectors on the right apart and collecting terms with like parameters,
Homogeneous system: (x1,x2,x3,x4,x5) = r(-3,1,0,0,0) + s(-4,0,-2,1,0,0) + t(-2,0,0,0,1,0)
Nonhomogeneous system: (x1,x2,x3,x4,x5) = r(-3,1,0,0,0) + s(-4,0,-2,1,0,0) + t(-2,0,0,0,1,0) + (0,0,0,0,0,1/3)
Each solution of the nonhomogeneous system can be obtained by adding (0,0,0,0,0,1/3) to the corresponding solution of the homogeneous system.

83. Theorem 3.4.4
The general solution of a consistent linear system Ax=b can be obtained by adding any specific solution of Ax=b to the general solution of Ax=0.
Proof:
Let x0 be any specific solution of Ax=b, Let W denote the solution set of Ax=0, and let x0+W denote the set of all vectors that result by adding x0 to each vector in W.
Show that if x is a vector in x0+W, then x is a solution of Ax=b, and conversely, that every solution of Ax=b is in the set x0+W.

84. Theorem 3.4.4
Assume that x is a vector in x0+W. This implies that x is expressible in the form x=x0+w, where Ax0=b and Aw=0. Thus, Ax = A(x0+w) = Ax0 + Aw = b + 0 = b which shows that x is a solution of Ax=b.
Conversely, let x be any solution of Ax=b. To show that x is in the set x0+W we must show that x is expressible in the form: x = x0+w, where w is in W (Aw = 0). We can do this by taking w = x-x0. It is in W since Aw = A(x-x0) = Ax – Ax0 = b – b = 0.

85. Geometric Interpretation of Theorem 3.4.4
We interpret vector addition as translation, then the theorem states that if x0 is any specific solution of Ax=b, then the entire solution set of Ax=b can be obtained by translating the solution set of Ax=0 by the vector x0.
Ax = b
Ax = 0 x0

86. 3.5 Cross Product

87. u×v = (u2v3 – u3v2, u3v1 – u1v3, u1v2 – u2v1)
Definition
If u = (u1,u2,u3) and v=(v1,v2,v3) are vectors in 3-space, then the cross product (外積) u×v is the vector defined by
u×v = (u2v3 – u3v2, u3v1 – u1v3, u1v2 – u2v1)
Or, in determinant notation
Remark: For the matrix to find the first component of u×v, delete the first column and take the determinant, …

88. Example
Find u×v, where u=(1,2,-2) and v=(3,0,1)
Solution

89. Theorems
Theorem (Relationships Involving Cross Product and Dot Product)
If u, v and w are vectors in 3-space, then
u · (u  v) = 0 (u  v is orthogonal to u)
v · (u  v) = 0 (u  v is orthogonal to v)
|| u  v ||2 = ||u||2||v||2 – (u · v)2 (Lagrange’s identity)
u  (v  w) = (u · w) v – (u · v) w (relationship between cross & dot product)
(u  v)  w = (u · w) v – (v · w) u (relationship between cross & dot product)

90. Proof of Theorem 3.5.1(a) Example: u=(1, 2, -2) and v=(3, 0, 1)
Let u=(u1,u2,u3) and v=(v1,v2,v3)
Example: u=(1, 2, -2) and v=(3, 0, 1)

91. Proof of Theorem 3.5.1(c)

92. Theorems Proof of (a) Theorem 3.5.2 (Properties of Cross Product)
If u, v and w are any vectors in 3-space and k is any scalar, then
u  v = - (v  u)
u  (v + w) = u  v + u  w
(u + v)  w = u  w + v  w
k(u  v) = (ku)  v = u  (kv)
u  0 = 0  u = 0
u  u = 0
Proof of (a)
Interchanging u and v interchanges the rows of the three determinants and hence changes the sign of each component in the cross product. Thus u  v = - (v  u).

93. Standard Unit Vectors The vectorsz
Standard Unit Vectors
k=(0,0,1)
j=(0,1,0) y
i=(1,0,0) x
The vectors
i = (1,0,0), j = (0,1,0), k = (0,0,1)
have length 1 and lie along the coordinate axes. They are called the standard unit vectors in 3-space.
Every vector v = (v1, v2, v3) in 3-space is expressible in terms of i, j, k since we can write
v = (v1, v2, v3) = v1(1,0,0) + v2 (0,1,0) + v3 (0,0,1) = v1i + v2j + v3k
For example, (2, -3, 4) = 2i – 3j +4k
Note that
i  i = 0, j  j = 0, k  k = 0
i  j = k, j  k = i, k  i = j
j  i = -k, k  j = -i, i  k = -j

94. Cross Product
A cross product can be represented symbolically in the form of 33 determinant:
Example: if u=(1,2,-2) and v=(3,0,1)

95. Cross Product It’s not true in general that For example:
Right-hand rule
If the fingers of the right hand are cupped so they point in the direction of rotation, then the thumb indicates the direction of u v

96. Geometric Interpretation of Cross Product
From Lagrange’s identity, we have
Since , it follows that so

97. Geometric Interpretation of Cross Product
From Lagrange’s identity in Theorem 3.5.1
If θ denotes the angle between u and v, then
Since , it follows that , thus

98. Geometric Interpretation of Cross Product
is the altitude (頂垂線) of the parallelogram determined by u and v. Thus, the area A of this parallelogram is given by
This result is even correct if u and v are collinear, since we have when

99. Area of a Parallelogram
Theorem (Area of a Parallelogram)
If u and v are vectors in 3-space, then ||u  v|| is equal to the area of the parallelogram determined by u and v.
Example
Find the area of the triangle determined by the point (2,2,0), (-1,0,2), and (0,4,3).

100. Triple Product Definition
If u, v and w are vectors in 3-space, then u · (v  w) is called the scalar triple product (純量三乘積) of u, v and w.

101. Example
u = 3i – 2j – 5k, v = i + 4j – 4k, w = 3j + 2k

102. Triple Product Remarks:
The symbol (u · v)  w make no sense because we cannot form the cross product of a scalar and a vector.
u · (v  w) = w · (u  v) = v · (w  u) , since the determinants that represent these products can be obtained from one another by two row interchanges.

103. Theorem 3.5.4
The absolute value of the determinant is equal to the area of the parallelogram in 2-space determined by the vectors u = (u1, u2), and v = (v1, v2),
The absolute value of the determinant is equal to the volume of the parallelepiped in 3-space determined by the vectors u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3),

104. Proof of Theorem 3.5.4(a) v u
View u and v as vectors in the xy-plane of an xyz-coordinate system. Express u=(u1,u2,0) and v=(v1,v2,0)
It follows from Theorem and the fact that that the area A of the parallelogram determined by u and v is

105. Proof of Theorem 3.5.4(b) The area of the base is
The height h of the parallelepiped is the length of the orthogonal projection of u on
The volume V of the parallelepiped is

106. Remark

107. Remark
We can conclude that where + or – results depending on whether u makes an acute or an obtuse angle with

108. Theorem 3.5.5
If the vectors u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3) have the same initial point, then they lie in the same plane if and only if